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table = '4E3A Single-core armoured 90 °C themosetting, copper conductors'; CSA = 50; % cross sectional area of cable, mm2 Length = 80; % length of cable, m % Uo = 400; % line-line voltage, V Ib = 152; % load current, A pf = 0.87; % power factor phases = 3; % number of phases (0 = d.c.) tp = 90; % maximum permitted cable temperature, °C
Ensure that we have both cos(ph) and sin(phi) from the power factor.
cos_phi = pf; sin_phi = sin(acos(pf)); disp(''); disp(['pf = cos(phi) = ', num2str(cos_phi), ' which gives sin(phi)= ', num2str(sin_phi, '%0.2f')]);
pf = cos(phi) = 0.87 which gives sin(phi)= 0.49
It = 222; % tablulated current A, table 4E3A, column 6 mVr = 0.86; % (mV/A/m)r table 4E3B, column 5 mVx = 0.29; % (mV/A/m)x table 4E3B, column 5 Ca = 0.96; % ambient temperature derating, table 4B1 Cg = 1.00; % grouping derating, [no grouping] Cs = 1.00; % soil thermal resistivyt derating [not buired] Cd = 1.00; % depth of burial derating [not buried]
Ct = (230+tp-(Ca^2*Cg^2*Cs^2*Cd^2-(Ib^2/It^2))*(tp-30))/(230+tp); disp(['Temperature correction factor Ct = ', num2str(Ct, '%0.2f' )]); vd = ( Ct*cos_phi*mVr + sin_phi*mVx ) * Ib * Length /1000; vdPercent = 100*vd/Uo; disp('Voltage drop = ', num2str(vd, '%0.2f'), ' V [', num2str(vdPercent, '%0.2f'), '%]');
Temperature correction factor Ct = 0.92 Voltage drop = 10.06 V [2.52%]
Published with MATLAB® R2015a